Roadmap. Deadlock Prevention. Deadlock Prevention (Cont.) Deadlock Detection. Exercise. Tevfik Koşar. CSE 421/521 - Operating Systems Fall 2012

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CSE 421/521 - Operating Systems Fall 2012 Roadmap Lecture - XI Deadlocks - II Deadlocks Deadlock Prevention Deadlock Detection Deadlock Recovery Deadlock Avoidance Tevfik Koşar University at Buffalo October 4 th, 2012 1 2 Deadlock Prevention! Ensure one of the deadlock conditions cannot hold!restrain the ways request can be made. Mutual Exclusion not required for sharable resources; must hold for nonsharable resources. Eg. read-only files Hold and Wait must guarantee that whenever a process requests a resource, it does not hold any other resources. 1. Require process to request and be allocated all its resources before it begins execution 2. or allow process to request resources only when the process has none. Example: Read from DVD to memory, then print. 1. holds printer unnecessarily for the entire execution Low resource utilization 2. may never get the printer later starvation possible Deadlock Prevention (Cont.) No Preemption If a process that is holding some resources requests another resource that cannot be immediately allocated to it, then all resources currently being held are released. Preempted resources are added to the list of resources for which the process is waiting. Process will be restarted only when it can regain its old resources, as well as the new ones that it is requesting. Circular Wait impose a total ordering of all resource types, and require that each process requests resources in an increasing order of enumeration. 3 4 Exercise Deadlock Detection (Silberschatz pp.249-251) to Allow system to enter deadlock state Detection algorithm Recovery scheme 5 6

Single Instance of Each Resource Type Maintain wait-for graph Nodes are processes. P i P j if P i is waiting for P j. Single Instance of Each Resource Type Periodically invoke an algorithm that searches for a cycle in the graph. An algorithm to detect a cycle in a graph requires an order of n 2 operations, where n is the number of vertices in the graph. Only good for single-instance resource allocation systems. Resource-Allocation Graph Corresponding wait-for graph 7 8 Several Instances of a Resource Type Detection Algorithm Available: A vector of length m indicates the number of available resources of each type. Allocation: An n x m matrix defines the number of resources of each type currently allocated to each process. Request: An n x m matrix indicates the current request of each process. If Request [i j ] = k, then process P i is requesting k more instances of resource type. R j. 1. Let Work and Finish be vectors of length m and n, respectively Initialize: (a) Work = Available (b) For i = 0,1, 2,, n-1, Finish[i] = false. 2. Find an index i such that both: (a) Finish[i] == false (b) Request i Work If no such i exists, go to step 4. 9 10 Detection Algorithm (Cont.) 3. Work = Work + Allocation i Finish[i] = true go to step 2. 4. If Finish[i] == false, for some i, 0 i n-1, then the system is in deadlock state. Moreover, if Finish[i] == false, then P i is deadlocked. Algorithm requires an order of O(m x n 2) operations to detect whether the system is in deadlocked state. 11 Example of Detection Algorithm Five processes P 0 through P 4 ; three resource types A (7 instances), B (2 instances), and C (6 instances). Allocation Request Available Work A B C P 0 0 1 0 0 0 0 0 0 0 0 0 0 P 1 2 0 0 2 0 2 P 2 3 0 3 0 0 0 P 3 2 1 1 1 0 0 P 4 0 0 2 0 0 2 Sequence <P 0, P 2, P 3, P 1, P 4 > will result in Finish[i] = true for all i. 12

Example (Cont.) Recovery from Deadlock: Process Termination P 2 requests an additional instance of type C. Allocation Request Available Work A B C P 0 0 1 0 0 0 0 0 0 0 0 0 0 P 1 2 0 0 2 0 2 P 2 3 0 3 0 0 1 P 3 2 1 1 1 0 0 P 4 0 0 2 0 0 2 State of system? Can reclaim resources held by process P 0, but insufficient resources to fulfill other processes; requests. Deadlock exists, consisting of processes P 1, P 2, P 3, and P 4. 13 Abort all deadlocked processes. --> expensive Abort one process at a time until the deadlock cycle is eliminated. --> overhead of deadlock detection alg. In which order should we choose to abort? Priority of the process. How long process has computed, and how much longer to completion. Resources the process has used. Resources process needs to complete. How many processes will need to be terminated. Is process interactive or batch? 24 Recovery from Deadlock: Resource Preemption Selecting a victim minimize cost. Rollback return to some safe state, restart process for that state. Starvation same process may always be picked as victim, include number of rollback in cost factor. Deadlock Avoidance Deadlock Prevention: prevent deadlocks by restraining resources and making sure one of 4 necessary conditions for a deadlock does not hold. (system design) --> possible side effect: low device utilization and reduced system throughput Deadlock Avoidance: Requires that the system has some additional a priori information available. (dynamic request check) i.e. request disk and then printer.. or request at most n resources --> allows more concurrency Similar to the difference between a traffic light and a police officer directing the traffic! 25 3 Deadlock Avoidance Simplest and most useful model requires that each process declare the maximum number of resources of each type that it may need. The deadlock-avoidance algorithm dynamically examines the resource-allocation state to ensure that there can never be a circular-wait condition. Resource-allocation state is defined by the number of available and allocated resources, and the maximum demands of the processes. Safe State A state is safe if the system can allocate resources to each process (upto its maximum) in some order and can still avoid a deadlock. When a process requests an available resource, system must decide if immediate allocation leaves the system in a safe state. System is in safe state if there exists a safe sequence of all processes. 4 6

Safe State Sequence <P 1, P 2,, P n > is safe if for each P i, the resources that P i can still request can be satisfied by currently available resources + resources held by all the P j, with j<i. If P i resource needs are not immediately available, then P i can wait until all P j have finished. When P j is finished, P i can obtain needed resources, execute, return allocated resources, and terminate. When P i terminates, P i+1 can obtain its needed resources, and so on. If no such sequence exists, the state is unsafe! Example of Safe State Five processes P 0 through P 4 ; three resource types A (7 instances), B (2 instances), and C (6 instances). Allocation Request Available Work A B C P 0 0 1 0 0 0 0 0 0 0 0 0 0 P 1 2 0 0 2 0 2 P 2 3 0 3 0 0 0 P 3 2 1 1 1 0 0 P 4 0 0 2 0 0 2 Sequence <P 0, P 2, P 3, P 1, P 4 > represents a safe state 7 20 Basic Facts Safe, Unsafe, Deadlock State If a system is in safe state no deadlocks. If a system is in unsafe state possibility of deadlock. Avoidance ensure that a system will never enter an unsafe state. 8 9 Example Consider a system with 3 processes and 12 disks. At t = t0; Maximum Needs Current Allocation P1 10 5 P2 4 2 P3 9 2 Example (cont.) Consider a system with 3 processes and 12 disks. At t = t1; Maximum Needs Current Allocation P1 10 5 P2 4 2 P3 9 3 23 24

Resource-Allocation Graph Algorithm Resource-Allocation Graph For Deadlock Avoidance Claim edge P i R j indicated that process P j may request resource R j ; represented by a dashed line. Claim edge converts to request edge when a process requests a resource. When a resource is released by a process, assignment edge reconverts to a claim edge. Resources must be claimed a priori in the system. 12 13 Unsafe State In Resource-Allocation Graph Banker s Algorithm Works for multiple resource instances. Each process declares maximum # of resources it may need. When a process requests a resource, it may have to wait if this leads to an unsafe state. When a process gets all its resources it must return them in a finite amount of time. 14 15 Data Structures for the Banker s Algorithm Let n = number of processes, and m = number of resources types. Available: Vector of length m. If available [j] = k, there are k instances of resource type R j available. Max: n x m matrix. If Max [i,j] = k, then process P i may request at most k instances of resource type R j. Allocation: n x m matrix. If Allocation[i,j] = k then P i is currently allocated k instances of R j. Need: n x m matrix. If Need[i,j] = k, then P i may need k more instances of R j to complete its task. Safety Algorithm 1. Let Work and Finish be vectors of length m and n, respectively. Initialize: Work = Available Finish [i] = false for i = 1,2,, n. 2. Find an i such that both: (a) Finish [i] = false (b) Need i Work If no such i exists, go to step 4. 3. Work = Work + Allocation i Finish[i] = true go to step 2. 4. If Finish [i] == true for all i, then the system is in a safe state. Need [i,j] = Max[i,j] Allocation [i,j]. 16 17

Resource-Request Algorithm for Process P i Let Request i be the request vector for process P i. If Request i [j] = k then process P i wants k instances of resource type R j. 1. If Request i Need i go to step 2. Otherwise, raise error condition, since process has exceeded its maximum claim. 2. If Request i Available, go to step 3. Otherwise P i must wait, since resources are not available. 3. Pretend to allocate requested resources to P i by modifying the state as follows: Available = Available - Request i ; Allocation i = Allocation i + Request i ; Need i = Need i Request i ; If safe the resources are allocated to Pi. If unsafe Pi must wait, and the old resource-allocation state is restored 5 processes P 0 through P 4 ; 3 resource types: A (10 instances), B (5 instances), and C (7 instances). Allocation Max Available P 0 0 1 0 7 5 3 3 3 2 P 1 2 0 0 3 2 2 P 2 3 0 2 9 0 2 P 3 2 1 1 2 2 2 P 4 0 0 2 4 3 3 18 19 The content of the matrix. Need is defined to be Max Allocation. Need A B C P 0 7 4 3 P 1 1 2 2 P 2 6 0 0 P 3 0 1 1 P 4 4 3 1 Allocation Max Available P 0 0 1 0 7 5 3 3 3 2 P 1 2 0 0 3 2 2 P 2 3 0 2 9 0 2 P 3 2 1 1 2 2 2 P 4 0 0 2 4 3 3 Need A B C 7 4 3 1 2 2 6 0 0 0 1 1 4 3 1 20 21 Allocation Max Available P 0 0 1 0 7 5 3 3 3 2 P 1 2 0 0 3 2 2 P 2 3 0 2 9 0 2 P 3 2 1 1 2 2 2 P 4 0 0 2 4 3 3 Need A B C 7 4 3 1 2 2 6 0 0 0 1 1 4 3 1 The system is in a safe state since the sequence < P 1, P 3, P 4, P 2, P 0 > satisfies safety criteria. 22 Example: P 1 Requests (1,0,2) Check that Request Available (that is, (1,0,2) (3,3,2) true. Allocation Need Available P 0 0 1 0 7 4 3 2 3 0 P 1 3 0 2 0 2 0 P 2 3 0 1 6 0 0 P 3 2 1 1 0 1 1 P 4 0 0 2 4 3 1 Executing safety algorithm shows that sequence <P1, P3, P4, P0, P2> satisfies safety requirement. Can request for (3,3,0) by P4 be granted? Can request for (0,2,0) by P0 be granted? 23

Summary Acknowledgements Deadlocks Deadlock Prevention Deadlock Detection Deadlock Recovery Deadlock Avoidance Hmm.. Operating Systems Concepts book and supplementary material by A. Silberschatz, P. Galvin and G. Gagne Operating Systems: Internals and Design Principles book and supplementary material by W. Stallings Modern Operating Systems book and supplementary material by A. Tanenbaum R. Doursat and M. Yuksel from UNR 37 38